Question

if two circle intersect at two points , prove that there Central lie on the perpendicular bisector of the common chord​

Answer 1

Answer:

Let the two circles with centre P and Q intersect at points A and B.

Join AB. AB is the common chord.

Join PQ. AB and PQ bisect each other at M.

Let M be the midpoint of AB.

Hence, PM ⊥ AB [Since, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord]

⇒ ∠PMA = 90º

Now, since M is the midpoint of AB

Hence, QM ⊥ AB

⇒ ∠QMA = 90º

Thus, ∠PMQ = ∠PMA + ∠QMA = 90º + 90º = 180º

Hence PMQ is a straight line and PMQ ⊥ AB

Therefore, PMQ is the perpendicular bisector of the common chord AB and passes through the two centers P and Q.

So, the centres lie on the perpendicular bisector of the common chords.

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Step-by-step explanation:

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Answer 2

The Answer is in Attachment

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